Water as leaving group

The reaction starts of with a protonation - use the catalyst. Resist the urge to protonate the 4-hydroxyl, but go for the one at position 1 that has the added functionality of the hemiacetal linkage. It is going to be the more reactive one. Protonation is followed by loss of water as leaving group. The intermediate oxonium cation shown is actually a resonance form of the simpler carbocation now you can see the role of the adjacent oxygen. The reaction is completed by attack of the nucleophile, the 4-hydroxyl of another molecule. This is not special, but is merely another version of the hemiacetal synthesis done in part (a). 

The principal difference hes in the poorer ability of amide ions to act as leaving groups, compared to alkoxides. As a result, protonation at nitrogen is required for breakdown of the tetrahedral intermediate. Also, exchange between the carbonyl oxygen and water is extensive because reversal of the tetrahedral intermediate to reactants is faster than its decomposition to products. 

A further explanation may lie in the superior properties of anilines as leaving groups when compared with water. This idea has, however, been disputed11,12 and a complex reaction involving all four components of the system suggested. 

Finally, HOH and ROH can leave protonated alcohols or ethers as leaving groups, hut neither the OH group (from alcohols) nor the OR group (from ethers, except for epoxides, see above) can leave. Remember, the conjugate acid of water is the hydronium ion, a strong acid (pKa = -1.7)... 

We have mostly seen halides and water from protonated alcohols as leaving groups in both S l and Sn2 reactions. Now we need to establish the principles that make for good and bad leaving groups. We might be considering an SnI reaction. 

If the A-aryl group is strongly activated, then it can be removed in nucleophilic substitution reactions in which the azole anion acts as leaving group. Thus l-(2,4-dinitrophenyl)-pyrazole and -imidazole react with N2H4 or NaOMe. Their A-picryl derivatives react similarly with BuNIT or water at a range of pH values. 

In acid solution the first reaction is similar, though the tetrahedral intermediate is neutral, and the carboxyl is still the better leaving group. The second esterification is now all right because methanol can attack the protonated carboxyhc acid and water can be driven out after a second protonation. The second step is an equihbrium, with water and methanol about equal as leaving groups, but methanol is present in large excess as the solvent and drives the equihbrium across. We have omitted proton transfer steps. 

When a hydroxy bridge is formed between two metal centers by nucleophilic substitution reaction (where nucleophile is hydroxyl group and water is leaving group), then a condensation process is called as olation [113]. 

An intramolecular Mizoroki-Heck reaction was carried out with Pd(OAc)2 (10 mol%) in the presence of P(o-tol)3 in an acetonitrile-water mixture [182], while another group used the same solvent mixture in an intramolecular Mizoroki-Heck reaction with alkoxy groups as leaving groups [183, 184]. The reaction was conducted at room temperature in the presence of Pd(OAc)2 (10 mol%) and a variety of phosphine ligands tpp, P(o-tol)3, P(p-tol)3, P(2-furyl)3, diphenylphosphinopropane, diphenylphosphinobutane and diphenylphosphi-noferrocene. 

The mechanism may be viewed as a series of nucleophilic attacks by water on the P+ centers with the 0x0 bridges in the P-O-P linkages as leaving groups, as shown below ... 

Hydrolysis of this PMP-a-keto acid imine in step 4 then completes the first part of the deamination reaction. The hydrolysis is the exact mechanistic reverse of imine formation (see Figure 14.8) and occurs hy nucleophilic addition of water to the imine, followed by proton transfer and expulsion of PMP as leaving group. 

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